If you force a very high current (much higher than the short-circuit current) into the battery's negative node, and out of the positive node, you'll be able to remove all the energy in it in a minimum time, but that time cannot be zero.
Also, to be able to inject that current, you'll need to treat the battery as a load, and connect it, upside down, to a high voltage source or (preferably) a high current source. The battery won't like it, though.
The internal resistance of a battery does not limit the current. That would be true only if all the voltage that we had access to was the one from the battery itself. But that is not true. No one impedes us from connecting an external voltage. The question asks for the fastest way to remove all possible energy from a battery.
In the following example, the short-circuit current of the battery would be 1.5/0.2=7.5 A. By connecting it upside down to a voltage generator with (for instance) 100 V and 0.01 ohm, we can force a current of (100+1.5)/(0.2+0.01)=483 A (in that ideal model), which is much higher than the short circuit current.
If you slowly transfer the energy in a battery to a supercap, and then quickly consume the energy in the supercap, that is not answering the question. The question asks how can we use all the energy in a battery (not in a cap) in an instant.